The New Sudoku Players' Forum

[SpAce]

The first thing that jumps out to my eye, even without candidates, is the Unique Rectangle (19)r13c29. We've talked about those before. It solves r3c2, after which it's singles.

Without candidates, i would say, it solves r3c3:

That works too. This is about the easiest possible UR configuration (3/4 cells obviously locked with the UR candidates + digit 9 just as obviously locked in all UR cells) so there are many possibilities to solve it. I tend to use UR Type 1 logic when it's available (i.e. place r3c2 in this case), but just as well you can use UR Type 4 logic and place r3c3, or you could place r1c2 just as easily. All are trivial with or without candidates, if you know what you're doing. UR Type 1 is probably the easiest to grasp, so I'd recommend starting with that if UR concepts are new.

There is also a naked pair in row 5, which - also without candidates - is easier to spot for me than the hidden triple in that row.

It's a good row to practice spotting various locked sets, as it has a degenerate naked quad/hidden quad pairing at the top level. Because it's degenerate, the naked quad contains a naked triple which contains a naked pair (so you get two singles when you break it). Similarly the hidden quad contains the hidden triple you mentioned (as long as 9r5c9 has been eliminated).

[eleven]

It's not type 1 or 4 logic, but internal/external candidates logic.
As without candidates hidden subsets are easier to spot (normally, not in this case), for UR's it's easier to find cells, where a candidate must be than to look, which candidates can be eliminated.

[SpAce]

It's not type 1 or 4 logic, but internal/external candidates logic.

I don't think they're mutually exclusive concepts. Type 1 and 4 are just pieces of very simple pre-packaged and named internal/external logic. To me your technique sounded a lot like Type 4 logic even if you didn't think in those terms.

As without candidates hidden subsets are easier to spot (normally, not in this case), for UR's it's easier to find cells, where a candidate must be than to look, which candidates can be eliminated.

For you! You have to remember that my no-pm solving skills are way inferior and always will be. I've forced myself to learn a bit of it, but I still tend to look at things in terms of candidates and eliminations because it's much more natural for me. It probably limits my no-pm solving quite a bit but it's not a high-priority for me anyway. I'm mostly doing it at all to challenge my brain a bit differently and to gain some new perspectives (which has actually helped my overall solving skills), but I don't really enjoy it except with relatively easy puzzles.

Therefore, for me UR Type 1 / internal logic is clearly easiest in this case. Why wouldn't it be? The only cell capable of holding extra candidates (internals) is obviously r3c2 (because r1 and b3 are both locked), and there's only one possibility because b1 has only three unsolved digits (so it must be the one that is not 1 or 9, i.e. 5). That's UR Type 1, and I can't really think of anything easier, because the only information required (besides knowing the locked UR cells) is contained within box 1. Even in much more complicated UR scenarios I look at internals first and then, if need be, externals. Seems simpler that way.

UR Type 4 / external logic is almost as easy when you notice that 9s are locked in r13c2 so 1 can't be there (so it must be in r3c3), but that requires looking at both b1 and c3 (in addition to the UR). I think it's very similar to seeing 1r3c3 as an external that must be placed.

The other way to use UR Type 1 is to see that since 9 can't be in r3c2 or r3c3 it must be in r1c2, but like Type 4, it requires looking at three pieces of information.

[sudokudumb]

Sorry guys
I'm even more confused now
How do we know that 5 isn't in r3c3? Or why/how we know that 1 is in r3c3? I'm so confused

[eleven]

This pattern is not possible in a unique sudoku

Code: Select all

  .  19  . | .  . 19
  .   .  . | .  .  .
  .  19  . | .  . 19


because it would have 2 solutions

Code: Select all

  x   1  x | x  x  9        x   9  x | x  x  1
  x   x  x | x  x  x        x   x  x | x  x  x
  x   9  x | x  x  1        x   1  x | x  x  9


(swapping the digits keeps a 1 and 9 in the 2 rows, columns and boxes, so it would give a valid solution too)

In your puzzle you have

Code: Select all

  x   *  x | x  x 19
  x   x  x | x  x  x
  x   *  . | x  x 19
---------------------
  x   x  x
  x   .  .
  x   .  9

So the cells marked * must not be 19. One of the digits must be elsewhere in the box. The only free place is r3c3. It cannot be 9, because there is a 9 in r6c3. So r3c3 must be 1.
(also without the 9r6c3 you would know, that 1 or 9 must be in r3c3 - in this case 5 could not be there and has to be in the * cells).

[Added:] This is a long explanation. Once you are familiar with it, you would fill in the 1 in 2 seconds. Such situations are not very common, but i find them again and again in newspaper puzzles, though the creators are not aware of them. However they seldom solve the puzzle like here.

To get the pencilmark explanations, just read about unique rectangles type 1 (or 4).

[SpAce]

Eleven already gave an excellent explanation and his method to see it without pencil marks, so here's one with them:

Code: Select all

.--------------.-------------.-----------------.
| 3  *19    6  | 5   7   4   | 2     8    *19  |
| 4   2     8  | 1   3   9   | 7     6     5   |
| 7  *19+5  15 | 8   6   2   | 4     3    *19  |
:--------------+-------------+-----------------:
| 8   4     3  | 29  25  7   | 59    1     6   |
| 2   157   15 | 69  15  136 | 3589  457   489 |
| 6   157   9  | 4   8   13  | 35    57    2   |
:--------------+-------------+-----------------:
| 1   8     4  | 3   9   5   | 6     2     7   |
| 5   3     7  | 26  24  68  | 1     9     48  |
| 9   6     2  | 7   14  18  | 58    45    3   |
'--------------'-------------'-----------------'

The UR cells are marked with the *. Notice that three of them (r1c2, r1c9, r3c9) can only be 1 or 9, so the only way to avoid the impossible rectangle (explained by eleven) is to have some other digit in the fourth corner (r3c2). It currently has three candidates (159), but because of the UR pattern we know that 1 and 9 aren't possible, so it must be 5. That's UR Type 1 logic, which is the simplest UR type. In that pattern three corners of the rectangle have only the UR digits as candidates, so they can be eliminated from the fourth. Those eliminations are always valid even if there's more than one extra candidate and you can't get an immediate placement.

In this case we also have a UR Type 4 pattern available. Its logic is a bit more complicated. In that you have to notice (just like with eleven's method) that the 9s are locked in the r13c2 cells, which means that 1 isn't possible there (or the other would be 9, and you'd have the impossible pattern). So, the two 1s can be eliminated from r13c2, which means you can place 1 in r3c3 (the only place left in the box). Again, even if you can't get an immediate placement, those eliminations are still valid.

More about this stuff here, here, and here.

[SpAce]

Here's a generic UR Type 1 (only):

Code: Select all

  x   ab   x   | x  x  ab
  x   x    x   | x  x  x
  x   abc  abc | x  x  ab
-------------------------
  x   x    x   |
  x   .    .   |
  x   .    .   |

   ||
   \/

  x   ab    x   | x  x  ab
  x   x     x   | x  x  x
  x   c-ab  abc | x  x  ab
---------------------------
  x   x     x   |
  x   .     .   |
  x   .     .   |

   ||
   \/

  x   ab    x    | x  x  ab
  x   x     x    | x  x  x
  x   c     ab   | x  x  ab
-------------------------
  x   x     x    |
  x   .     .    |
  x   .     .    |

Here's UR Type 4 + UR Type 1 (i.e. the situation we actually had):

Code: Select all

  x   ab    x   | x  x  ab
  x   x     x   | x  x  x
  x   abc   ac  | x  x  ab
-------------------------
  x   x     x   |
  x   .     .   |
  x   .     b   |

   ||
   \/

  x   b-a   x   | x  x  ab
  x   x     x   | x  x  x
  x   bc-a  ac  | x  x  ab
-------------------------
  x   x     x   |
  x   .     .   |
  x   .     b   |

   ||
   \/

  x   b    x   | x  x  a 
  x   x    x   | x  x  x
  x   c    a   | x  x  b
-------------------------
  x   x    x   |
  x   .    .   |
  x   .    b   |

Notice how it solves the whole pattern. It doesn't really matter which way you work it out (my diagram starts with UR Type 4 logic, but you could start with Type 1 just as well). However, I'm beginning to believe that eleven's way of solving it is probably fastest without pencil marks (though not by a large margin). In my experience these situations are quite common in newspaper puzzles (like eleven said), so it might not be a bad idea to learn it well and use the fastest possible way of filling in those five cells.

[sudokudumb]

WOW
I have to learn so much more
Thanks guys, without your directions I will be lost